I am tryting to split string by "," using StringTokenizer but not able to get whole values , token count shows 3 but printing only two elements, i have added my code below
if i have tried with two other inputs
"Ravi,Tuti,786" - same output
"Ravi,Tuti,786,pincode" getting three tokens not last one
public class Tokenizer{
public static void main(String[] args){
String str = "Ravi,Tuti,786";//survival of fittest,journey to get job,update skill,try,get job";
StringTokenizer stk = new StringTokenizer(str,",");
System.out.println(stk.countTokens());
for(int i=0;i<=stk.countTokens();i++){
System.out.println(stk.nextToken());}
}
}
output is
3
Ravi
Tuti
Use hasMoreTokens() with nextToken:
public class Tokenizer{
public static void main(String[] args){
String str = "Ravi,Tuti,786";//survival of fittest,journey to get job,update skill,try,get job";
StringTokenizer stk = new StringTokenizer(str,",");
System.out.println(stk.countTokens());
while (stk.hasMoreTokens()) {
System.out.println(stk.nextToken());
}
}
}
The problem with your approach is that you are running countTokens in the for loop, which changes after nextToken is called.
If you want to use a for loop, you need to save the token count to a variable:
int numTokens = stk.countTokens();
for (int i = 0; i < numTokens; i++) {
System.out.println(stk.nextToken());
}
You should use hasTokens() method.
for( ; stk.hasMoreTokens() ; ) {
System.out.println(stk.nextToken());
}
The countTokens() method returns:
the number of tokens remaining in the string using the current delimiter set.
So in your if loop it keeps getting evaluating and returning smaller numbers. To prevent this you can resolve it to a variable
int length = stk.countTokens();
for(int i=0;i<length;i++){
System.out.println(stk.nextToken());
}
If you do not wish to introduce another variable you can start i at what countTokens() returns and then loop until i is more that zero (While subtracting from i instead of adding)
for(int i=stk.countTokens();i>0;i--){
System.out.println(stk.nextToken());
}
Output:
3
Ravi
Tuti
786
Related
public class sortingtext {
public static void main(String[] args) throws IOException {
String readline="i have a sentence with words";
String[] words=readline.split(" ");
Arrays.sort(words, (a, b)->Integer.compare(b.length(), a.length()));
for (int i=0;i<words.length;i++)
{
int len = words[i].length();
int t=0;
System.out.println(len +"-"+words[i]);
}
}
input:
i have a sentence with words
My code split a string and then it should print each word and their length.
The output I get looks like:
8- sentence
5- words
4- have
4-with
1-I
1-a
I want to group the words of same length to get that:
8- sentence
5- words
4- have ,with
1- I ,a
But I don't get how to group them.
Easy with the stream API:
final Map<Integer, List<String>> lengthToWords = new TreeMap<>(
Arrays.stream(words)
.collect(Collectors.groupingBy(String::length))
);
The stream groups the words by length into a map (implementation detail, but it will be a HashMap), the TreeMap then sorts this map based on the key (the word length).
Alternatively, you can write it like this which is more efficient but in my opinion less readable.
final Map<Integer, List<String>> lengthToWords = Arrays.stream(words)
.collect(Collectors.groupingBy(String::length, TreeMap::new, Collectors.toList()));
If you are a beginner or not familiar with stream API:
public static void main(String[] args) {
String readline= "i have a sentence with words";
String[] words = readline.split(" ");
Arrays.sort(words, (a, b)->Integer.compare(b.length(), a.length()));
// declare a variable to hold the current string length
int currLength = -1;
for(int i = 0; i<words.length; i++){
if(currLength == words[i].length()){
// if currLength is equal to current word length just append a comma and this word
System.out.print(", "+words[i]);
}
else{
// if not update currLength, jump to a new line and print new length with the current word
currLength = words[i].length();
System.out.println();
System.out.print(currLength+ " - "+words[i]);
}
}
}
Note: The println("...") method prints the string "..." and moves the cursor to a new line. The print("...") method instead prints just the string "...", but does not move the cursor to a new line. Hence, subsequent printing instructions will print on the same line. The println() method can also be used without parameters, to position the cursor on the next line.
Good day, guys,
I'm working on a program which requires me to input a name (E.g Patrick-Connor-O'Neill). The name can be composed of as many names as possible, so not necessarily restricted to solely 3 as seen in the example above.But the point of the program is to return the initials back so in this case PCO. I'm writing to ask for a little clarification. I need to separate the names out from the hyphens first, right? Then I need to take the first character of the names and print that out?
Anyway, my question is basically how do I separate the string if I don't know how much is inputted? I get that if it's only like two terms I would do:
final String s = "Before-After";
final String before = s.split("-")[0]; // "Before"
I did attempt to do the code, and all I have so far is:
import java.util.Scanner;
public class main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String input = scan.nextLine();
String[] x = input.split("-");
int u =0;
for(String i : x) {
String y = input.split("-")[u];
u++;
}
}
}
I'm taking a crash course in programming, so easy concepts are hard for me.Thanks for reading!
You don't need to split it a second time. By doing String[] x = input.split("-"); you have an Array of Strings. Now you can iterate over them which you already do with the enhanced for loop. It should look like this
String[] x = input.split("-");
String initials = "";
for (String name : x) {
initials += name.charAt(0);
}
System.out.println(initials);
Here are some Java Docs for the used methods
String#split
String#charAt
Assignment operator +=
You can do it without splitting the string by using String.indexOf to find the next -; then just append the subsequent character to the initials:
String initials = "" + input.charAt(0);
int next = -1;
while (true) {
next = input.indexOf('-', next + 1);
if (next < 0) break;
initials += input.charAt(next + 1);
}
(There are lots of edge cases not handled here; omitted to get across the main point of the approach).
In your for-each loop append first character of all the elements of String array into an output String to get the initials:
String output = "";
for(String i : x) {
output = output + y.charAt(0);
}
This will help.
public static void main(String[] args) {
String output = "";
String input = "Patrick-Connor-O'Neil-Saint-Patricks-Day";
String[] brokenInput = input.split("-");
for (String temp : brokenInput) {
if (!temp.equals(""))
output = output + temp.charAt(0);
}
System.out.println(output);
}
You could totally try something like this (a little refactor of your code):
import java.util.Scanner;
public class main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String input = "";
System.out.println("What's your name?");
input = scan.nextLine();
String[] x = input.split("-");
int u =0;
for(String i : x) {
String y = input.split("-")[u];
u++;
System.out.println(y);
}
}
}
I think it's pretty easy and straightforward from here if you want to simply isolate the initials. If you are new to Java make sure you use a lot of System.out since it helps you a lot with debugging.
Good coding.
EDIT: You can use #Mohit Tyagi 's answer with mine to achieve the full thing if you are cheating :P
This might help
String test = "abs-bcd-cde-fgh-lik";
String[] splitArray = test.split("-");
StringBuffer stringBuffer = new StringBuffer();
for (int i = 0; i < splitArray.length; i++) {
stringBuffer.append(splitArray[i].charAt(0));
}
System.out.println(stringBuffer);
}
Using StringBuffer will save your memory as, if you use String a new object will get created every time you modify it.
I have to be able to input any two words as a string. Invoke a method that takes that string and returns the first word. Lastly display that word.
The method has to be a for loop method. I kind of know how to use substring, and I know how to return the first word by just using .substring(0,x) x being how long the first word is.
How can I make it so that no matter what phrase I use for the string, it will always return the first word? And please explain what you do, because this is my first year in a CS class. Thank you!
I have to be able to input any two words as a string
The zero, one, infinity design rule says there is no such thing as two. Lets design it to work with any number of words.
String words = "One two many lots"; // This will be our input
and then invoke and display the first word returned from the method,
So we need a method that takes a String and returns a String.
// Method that returns the first word
public static String firstWord(String input) {
return input.split(" ")[0]; // Create array of words and return the 0th word
}
static lets us call it from main without needing to create instances of anything. public lets us call it from another class if we want.
.split(" ") creates an array of Strings delimited at every space.
[0] indexes into that array and gives the first word since arrays in java are zero indexed (they start counting at 0).
and the method has to be a for loop method
Ah crap, then we have to do it the hard way.
// Method that returns the first word
public static String firstWord(String input) {
String result = ""; // Return empty string if no space found
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
result = input.substring(0, i);
break; // because we're done
}
}
return result;
}
I kind of know how to use substring, and I know how to return the first word by just using .substring(0,x) x being how long the first word is.
There it is, using those methods you mentioned and the for loop. What more could you want?
But how can I make it so that no matter what phrase I use for the string, it will always return the first word?
Man you're picky :) OK fine:
// Method that returns the first word
public static String firstWord(String input) {
String result = input; // if no space found later, input is the first word
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
result = input.substring(0, i);
break;
}
}
return result;
}
Put it all together it looks like this:
public class FirstWord {
public static void main(String[] args) throws Exception
{
String words = "One two many lots"; // This will be our input
System.out.println(firstWord(words));
}
// Method that returns the first word
public static String firstWord(String input) {
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
return input.substring(0, i);
}
}
return input;
}
}
And it prints this:
One
Hey wait, you changed the firstWord method there.
Yeah I did. This style avoids the need for a result string. Multiple returns are frowned on by old programmers that never got used to garbage collected languages or using finally. They want one place to clean up their resources but this is java so we don't care. Which style you should use depends on your instructor.
And please explain what you do, because this is my first year in a CS class. Thank you!
What do I do? I post awesome! :)
Hope it helps.
String line = "Hello my name is...";
int spaceIndex = line.indexOf(" ");
String firstWord = line.subString(0, spaceIndex);
So, you can think of line as an array of chars. Therefore, line.indexOf(" ") gets the index of the space in the line variable. Then, the substring part uses that information to get all of the characters leading up to spaceIndex. So, if space index is 5, it will the substring method will return the indexes of 0,1,2,3,4. This is therefore going to return your first word.
The first word is probably the substring that comes before the first space. So write:
int x = input.indexOf(" ");
But what if there is no space? x will be equal to -1, so you'll need to adjust it to the very end of the input:
if (x==-1) { x = input.length(); }
Then use that in your substring method, just as you were planning. Now you just have to handle the case where input is the blank string "", since there is no first word in that case.
Since you did not specify the order and what you consider as a word, I'll assume that you want to check in given sentence, until the first space.
Simply do
int indexOfSpace = sentence.indexOf(" ");
firstWord = indexOfSpace == -1 ? sentence : sentence.substring(0, indexOfSpace);
Note that this will give an IndexOutOfBoundException if there is no space in the sentence.
An alternative would be
String sentences[] = sentence.split(" ");
String firstWord = sentence[0];
Of if you really need a loop,
String firstWord = sentence;
for(int i = 0; i < sentence.length(); i++)
{
if(sentence.charAt(i) == ' ')
{
sentence = firstWord.substring(0, i);
break;
}
}
You may get the position of the 'space' character in the input string using String.indexOf(String str) which returns the index of the first occurrence of the string in passed to the method.
E.g.:
int spaceIndex = input.indexOf(" ");
String firstWord = input.substring(0, spaceIndex);
Maybe this can help you figure out the solution to your problem. Most users on this site don't like doing homework for students, before you ask a question, make sure to go over your ISC book examples. They're really helpful.
String Str = new String("Welcome to Stackoverflow");
System.out.print("Return Value :" );
System.out.println(Str.substring(5) );
System.out.print("Return Value :" );
System.out.println(Str.substring(5, 10) );
I created a method to output a String. Using the split method and a for loop, I added each word in my sentence into a String array, replacxing the last two letters of each word with "ed". Now, my return statement should return each of the words. When I used System.out.print, it worked. When I use a return and call it in my main method, I get this output: "[Ljava.lang.String;#1b6235b"
The error seems so simple but I just don't know where I'm going worng. Any help would be appreciated.
Here is my method:
public String[] processInfo() {
String sentence = this.phrase;
String[] words = sentence.split(" ");
if (!this.phrase.equalsIgnoreCase("Fred")) {
for (int i = 0; i < words.length; i++) {
words[i] = words[i].substring(0, words[i].length() - 2).concat(
"ed ");
// System.out.print(words[i]);
}
}
return words;
}
You are printing arrays but arrays don't have a proper implementation of toString() method by default.
What you see is
"[Ljava.lang.String;#1b6235b"
This is [Ljava.lang.String; is the name for String[].class, the java.lang.Class representing the class of array of String followed by its hashCode.
In order to print the array you should use Arrays.toString(..)
System.out.println(Arrays.toString(myArray));
A good idea however, it returns my Strings in an Array format. My aim
is to return them back into sentence format. So for example, if my
input is, "Hey my name is Fred", it would output as, "Hed ed naed ed
Fred". Sorry, I forgot to add that it also seperates it with commas
when using Arrays.toString
Then you should modify your processInfo() returning a String or creating a new method that convert your String[] to a String.
Example :
//you use like this
String [] processInfoArray = processInfo();
System.out.println(myToString(processInfoArray));
// and in another part you code something like this
public static String myToString(String[] array){
if(array == null || array.length == 0)
return "";
StringBuilder sb = new StringBuilder();
for(int i=0;i<array.length-1;i++){
sb.append(array[i]).append(" ");
}
return sb.append(array[array.length -1]).toString();
}
As much as I can get from your question and comment is that your aim is to return them back into sentence format. So for example, if your input is, "Hey my name is Fred", it would output as, "Hed ed naed ed Fred".
In that case you should return a String, and not an array. I have modified your method a bit to do so. Let me know if you wanted something else.
public String processInfo() {
String sentence = this.phrase;
String[] words = sentence.split(" ");
if (!this.phrase.equalsIgnoreCase("Fred")) {
sentence = "";
for (int i = 0; i < words.length; i++) {
words[i] = words[i].substring(0, words[i].length() - 2).concat(
"ed ");
sentence += " " + words[i];
// System.out.print(words[i]);
}
}
return sentence.trim();
}
Your commented out call to System.out.print is printing each element of the array from inside the loop. Your method is returning a String[]. When you try to print an array, you will get the java representation of the array as you are seeing. You either need to change your method to build and return a string with all the array entries concatenated together, or your calling code needs to loop through the returned array and print each entry.
I have a string (comma seperated)
For example:
a,bgf,sad,asd,rfw,fd,se,sdf,sdf,...
What I need is to extract a substring up to the 1000th comma.
How to achieve this in Java?
An efficient way of doing this would be to use indexof(int ch, int fromIndex) intead of using split(String regex, int limit) or split(String regex) especially if the given string is long.
This could be done something like this
[pseudocode]
asciiCodeForComma = 0x2c
nextIndex=0
loop 1000 times
nextIndex= csv.indexof(asciiCodeForComma , nextIndex)
requiredSubString = csv.subString(0, nextIndex)
String csv = "some,large,string";
String[] parts = csv.split(",");
String thousandthElement = null; // the default value if there are less than 1000
if (parts.length > 999)
thousandthElement = parts[999];
You can use StringTokenizer with comma as separator, then loop nextToken() 1000 times.
I think he is asking for all 1000 element.. This should solve your problem. Understand and then copy-paste as this is ur homework :)
public static void main(String[] args) {
// TODO Auto-generated method stub
String samplecsv = "a,bgf,sad,asd,rfw,fd,se,sdf,sdf,";
String[] splitedText = samplecsv.split(",");
StringBuffer newtext = new StringBuffer();
for (int i = 0; i < 3; i++) {
newtext.append(splitedText[i]);
}
System.out.println(newtext);
}
So to solve this problem what you need to do is understand how to extract a string from a delimiter separated input stream. Then executing this for the case for N strings is trivial. The pseudocode for doing this for the individual record is below:
function parse(inputValue, delimiter)
{
return inputValue.split(delimiter)
}
Now to do that for the case where there are N inputValues is equally as trivial.
function parse(inputValues, delimiter)
{
foreach inputValue in inputValues
returnValue.append(parse(inputValue,delimiter)
return returnValue
}
There is actually built in functionality to do this by putting a second argument into split. If what you're really looking for is the whole string before the 1000th comma, you can still use this function but you will have to concatenate the first section of the array.
public static void main(String[] args){
String sample = "s,a,m,p,l,e";
String[] splitSample = sample.split(",",1000);
if (splitSample.length == 1000)
System.out.println(splitSample[1000]);
else
System.out.println("Your error string");
}